The dplyr way would be: library(dplyr) df %>% group_by(col1, col2, col3) %>% summarise_each(funs(sum)) You can further specify the columns to be summarised or excluded from the summarise_each by using the special functions mentioned in the help file of ?dplyr::select.
Two ways to programmatically select variable(s): with = FALSE: DT = data.table(col1 = 1:3) colname = “col1” DT[, colname, with = FALSE] # col1 # 1: 1 # 2: 2 # 3: 3 ‘dot dot’ (..) prefix: DT[, ..colname] # col1 # 1: 1 # 2: 2 # 3: 3 For further description of the … Read more
Slightly more elegant: library(data.table) DT[ , .SD[which.min(Employees)], by = State] State Company Employees 1: AK D 24 2: RI E 19 Slighly less elegant than using .SD, but a bit faster (for data with many groups): DT[DT[ , .I[which.min(Employees)], by = State]$V1] Also, just replace the expression which.min(Employees) with Employees == min(Employees), if your data … Read more
Here’s the fast data.table way: bdt[bdt[, .I[g == max(g)], by = id]$V1] This avoids constructing .SD, which is the bottleneck in your expressions. edit: Actually, the main reason the OP is slow is not just that it has .SD in it, but the fact that it uses it in a particular way – by calling … Read more
Just a quick sum up. <- with data.table is just like base; i.e., no copy is taken until a subassign is done afterwards with <- (such as changing the column names or changing an element such as DT[i,j]<-v). Then it takes a copy of the whole object just like base. That’s known as copy-on-write. Would … Read more
You probably want to use the na.locf() function from the zoo package to carry the last observation forward to replace your NA values. Here is the beginning of its usage example from the help page: library(zoo) az <- zoo(1:6) bz <- zoo(c(2,NA,1,4,5,2)) na.locf(bz) 1 2 3 4 5 6 2 2 1 4 5 2 … Read more
Yes, in your formula, you can cbind the numeric variables to be aggregated: aggregate(cbind(x1, x2) ~ year + month, data = df1, sum, na.rm = TRUE) year month x1 x2 1 2000 1 7.862002 -7.469298 2 2001 1 276.758209 474.384252 3 2000 2 13.122369 -128.122613 … 23 2000 12 63.436507 449.794454 24 2001 12 999.472226 … Read more
If I understand OP correctly, you want smth like this: dt[, {bigN = .N; .SD[, .N / bigN, by = subg]}, by = group] or maybe (and very similarly) this: dt[, {counts.sum = sum(counts); .SD[, counts / counts.sum, by = subg]}, by = group]
I am going to try answering it myself. I am seeing Warning message in a hope to understand it better for any limitations. However, for me it works fine (ignoring the warning message). On a side note, data.table breaks all the limitation of R that stops it to do Big Data processing, and lest i … Read more
This seems to work: dt[ , (cols) := lapply(.SD, “*”, -1), .SDcols = cols] The result is a b d 1: -1 -1 1 2: -2 -2 2 3: -3 -3 3 There are a few tricks here: Because there are parentheses in (cols) :=, the result is assigned to the columns specified in cols, … Read more